Question

if the radius of sphere is increased by 50% then the ratio of percentage increase in volume to the percentage increase in surface area

Answer 1

Answer on Question #48820 – Math – Geometry

if the radius of sphere is increased by $50\%$ then the ratio of percentage increase in volume to the percentage increase in surface area

Solution:

Let the initial radius of the sphere $R_1 = R$ and final radius $R_2 = 1.5R$

Then, initial volume of the sphere: $V_{1} = \frac{4}{3}\pi R_{1}^{3} = \frac{4}{3}\pi R^{3}$

Final volume of the sphere: $V_{2} = \frac{4}{3}\pi R_{2}^{3} = \frac{4}{3}\pi (1.5R)^{3} = 1.5^{3}\cdot \frac{4}{3}\pi R^{3}$

Initial surface area of the sphere:

$$S _ {1} = 4 \pi R _ {1} ^ {2} = 4 \pi R ^ {2}$$

Final surface area of the sphere:

$$S _ {2} = 4 \pi R _ {2} ^ {2} = 4 \pi (1. 5 R) ^ {2} = 1. 5 ^ {2} \cdot 4 \pi R ^ {2}$$

Ratio of percentage increase in volume to the percentage increase in surface area

$$\frac {\text {increase in volume}}{\text {increase in surface}} = \frac {\frac {V _ {2}}{V _ {1}} - 1}{\frac {S _ {2}}{S _ {1}} - 1} = \frac {\frac {1 . 5 ^ {3} \cdot \frac {4}{3} \pi R ^ {3}}{4 \pi R ^ {2}} - 1}{\frac {1 . 5 ^ {2} \cdot 4 \pi R ^ {2}}{4 \pi R ^ {2}} - 1} = \frac {1 . 5 ^ {3} - 1}{1 . 5 ^ {2} - 1} = 1. 9$$

Answer: 1.9

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